Question

If $\alpha ,\beta ,\gamma$ are the roots of the equation $2x^3-3x^2+6x+1=0$, then ${\alpha }^2+{\beta }^2+{\gamma }^2$ is equal to

• A. $-\frac{15}{4}$
• B. $\frac{15}{4}$
• C. $\frac{9}{4}$
• D. $4$

Answer 1

Answer: (A)

$-\frac{15}{4}$

Given equation $2x^3-3x^2+6x+1=0$,
Sum of roots:

$\alpha +\beta +\gamma =\frac{-b}{a}$

So, $\alpha +\beta +\gamma =\frac{3}{2}$,

Product of roots:

$\alpha \beta \gamma =\frac{-d}{a}$

$\alpha \beta \gamma =\frac{-1}{2}$,

Sum of products taken $2$ at a time:

$\Sigma \alpha \beta =\alpha \beta +\beta \gamma +\alpha \gamma =\frac{c}{a}$

$\alpha \beta +\beta \gamma +\alpha \gamma =\frac{6}{2}=3$

Now, we know that

$(\alpha +\beta +\gamma {)}^2={\alpha }^2+{\beta }^2+{\gamma }^2+2(\alpha \beta +\beta \gamma +\gamma \alpha )$

Re-arranging, we get

${\alpha }^2+{\beta }^2+{\gamma }^2=(\alpha +\beta +\gamma {)}^2-2(\Sigma \alpha \beta )$
${\alpha }^2+{\beta }^2+{\gamma }^2={\left(\frac{3}{2}\right)}^2-2\times 3$

$=\frac{9}{4}-6$

$=-\frac{15}{4}$

Hence, option A.